Sufficiency for Truncated Geometric

Question

Here is a deviant of a question I feel like I have seen several times on truncated exponentials and similar distributions for finding sufficient statistics:

Let $$\mathbb{P}(Y=y)=\theta^y(1-\theta)^{\mathbb{I}\{y\leq z\}}$$ for some $z\in\mathbb{N}$ and $y\in\mathbb{N}$. Is the MLE for $\hat{\theta}$, $\frac{n\bar{Y}-n}{n-\sum\limits_{i=1}^n \mathbb{I}\{y=z+1\}}$, sufficient? If not, what else aside from the entire data $Y=(y_1,\dots,y_n)$, would constitute a sufficient statistic?

My Attempt: I believe that $T(Y)=\left(\sum y_i,\sum \mathbb{I}\{y=z+1\})\right)$ would easily work as a sufficient statistic. The MLE is a function (not 1-1 though) of this statistic which checks the box. As for if the MLE itself is sufficient (which intuitively I don't think is), we have

$$\mathcal{L}(\theta)=\theta^{\sum y_i}(1-\theta)^{n-\sum \mathbb{I}\{y_i=z+1\}},$$

but by inspection, I really see no way of (successfuly) using the factorization theorem here nor anything dealing with the exponential family as this does not reside in that family of distributions.

Question: Is there a more rigorous or correct way of going about this? I am being very hand-wavy here and was wondering if there's an alternative?

Answer 1

First, this truncated geometric pmf is not correct, it misses a factor so that it sums to one. (But that does not influence the algebra here). But for completeness, the correct formula must be $$ \mathbb{P}(Y=y) = \left( 1+\frac{\theta^{z+2}}{1-\theta} \right)^{-1} \theta^y(1-\theta)^{\mathbb{I}\{y\leq z\}} $$ (I do this as the pmf is written, since you call it a truncayed geometric, it might be that it is written erroneously, if, so, edit to correct)

Then we can write the likelihood, and use the factorization theorem: $$ \mathcal{L}=\prod_1^n \cdot \left( 1+\frac{\theta^{z+2}}{1-\theta} \right)^{-1} \cdot \theta^{y_i} \cdot (1-\theta)^{\mathbb{I}(y_i\le z)} \\ \propto \theta^{\sum y_i} \cdot (1-\theta)^{\sum_i \mathbb{I}(y_i\le z)} $$ and from this you can conclude!